problem A. Vasily the Bear and Triangle
题目：http://codeforces.com/problemset/problem/336/A
思路：水题，，虽然没大看懂什么意思，但随便写了写就过了。。。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 7 int main() 8 { 9 int x,y; 10 scanf("%d%d",&x,&y); 11 int ax,ay; 12 ax=x; 13 if(x<0) 14 ax=-x; 15 ay=y; 16 if(y<0) 17 ay=-y; 18 int s=ax+ay; 19 if(x<0) 20 { 21 if(y<0) 22 printf("%d 0 0 %d ",-s,-s); 23 else 24 { 25 printf("%d 0 0 %d ",-s,s); 26 } 27 } 28 else 29 { 30 if(y<0) 31 printf("0 %d %d 0 ",-s,s); 32 else 33 { 34 printf("0 %d %d 0 ",s,s); 35 } 36 } 37 return 0; 38 }
problem B. Vasily the Bear and Fly
题目：http://codeforces.com/problemset/problem/336/B
思路：从1~m号圆圆心分别到m+1~2m号圆圆心的最短距离的平均值
计算竖列相差i个圆 计算次数 每次距离
i=0 m 2R
i=1 2m-2 2R+sqrt(2)*R
i>=2 2m-2i 2*i-2+sqrt（8）*R
次数乘距离相加除以m*m即可
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 int main() 7 { 8 int m,r; 9 scanf("%d%d",&m,&r); 10 double dm=m; 11 double dr=r; 12 double sum; 13 int i; 14 double p=sqrt(2.0)*dr; 15 double pp=sqrt(8.0)*dr; 16 sum=dm*2*dr; 17 for(i=1; i<m; i++) 18 { 19 int u=2*i; 20 if(i<2) 21 sum+=(2*dm-u)*(u*dr+p); 22 else 23 sum+=(2*dm-u)*((u-2)*dr+pp); 24 } 25 sum/=(dm*dm); 26 printf("%.10f ",sum); 27 28 return 0; 29 }
problem C. Vasily the Bear and Sequence
题目：http://codeforces.com/contest/336/problem/C
思路：从大到小枚举v，若data[j]&1<<i!=0,,,则证明满足条件 否则不行，，，得到的sum若能整除1<<i,,,则找到。。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int maxn=100010; 6 int data[maxn]; 7 int main() 8 { 9 int n; 10 while(scanf("%d",&n)!=EOF) 11 { 12 int i,j; 13 for(i=1; i<=n; i++) 14 { 15 scanf("%d",&data[i]); 16 } 17 int sum; 18 int num; 19 for(i=29; i>=0; i--) 20 { 21 int m=1<<i; 22 sum=-1; 23 num=0; 24 for(j=1; j<=n; j++) 25 { 26 if((data[j]&m)!=0) 27 { 28 if(sum==-1) 29 sum=data[j]; 30 else 31 sum&=data[j]; 32 num++; 33 } 34 } 35 int k=0; 36 if(sum%m==0) 37 { 38 printf("%d ",num); 39 for(i=1;i<=n;i++) 40 { 41 if((data[i]&m)!=0) 42 { 43 k++; 44 if(k!=num) 45 printf("%d ",data[i]); 46 else 47 printf("%d ",data[i]); 48 } 49 50 } 51 break; 52 } 53 } 54 } 55 return 0; 56 }
problem D. Vasily the Bear and Beautiful Strings
题目：http://codeforces.com/contest/336/problem/D
思路：g值=string
0 =1__(指第一位只要为1，后面随便排列最后即可得到g值为0,(下同))
1 =01__
0 =001__
1 =0001__
......=........
即可发现规律，，，，注意1的个数只有1的时候需要特判
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 #define LL __int64 7 const int maxn=100010; 8 const int mod=1000000007; 9 int n0,n1,g; 10 LL fac[maxn*2]; 11 void init()//处理i的阶乘 12 { 13 int i; 14 fac[1]=1; 15 for(i=2; i<=maxn*2-5; i++) 16 { 17 fac[i]=i*fac[i-1]; 18 fac[i]%=mod; 19 } 20 } 21 //求p^n 22 LL poow(LL p,LL n) 23 { 24 LL sp=1; 25 while(n>0) 26 { 27 if(n%2==1) 28 sp=sp*p%mod; 29 n/=2; 30 p=p*p%mod; 31 } 32 return sp; 33 } 34 //在a与mod互质的情况，根据费马小定理，求a的逆元 35 LL inv(LL a) 36 { 37 return poow(a,mod-2); 38 } 39 //求组合数 40 LL c(int n,int k) 41 { 42 LL res; 43 if(n==k) 44 return 1; 45 if(k==0) 46 return 1; 47 res=fac[n]*inv(fac[n-k]*fac[k]%mod); 48 res%=mod; 49 return res; 50 } 51 int main() 52 { 53 init(); 54 while(scanf("%d%d%d",&n0,&n1,&g)!=EOF) 55 { 56 int num1,num0; 57 int i; 58 LL ans=0; 59 if(n0==0) 60 { 61 if(n1==1) 62 { 63 if(g==0) 64 printf("0 "); 65 else 66 printf("1 "); 67 } 68 else 69 { 70 if(g==0) 71 printf("1 "); 72 else 73 printf("0 "); 74 } 75 continue; 76 } 77 if(n1==0) 78 { 79 if(n0%2==1) 80 { 81 if(g==0) 82 printf("1 "); 83 else 84 printf("0 "); 85 } 86 else 87 { 88 if(g==0) 89 printf("0 "); 90 else 91 printf("1 "); 92 } 93 continue; 94 } 95 if(g==0) 96 { 97 for(i=0; i<=n0; i+=2)//偶数个0加上1个1时候，后面随便排列 98 { 99 num1=1; 100 num0=i; 101 if(n0+n1-(num0+num1)>0) 102 { 103 ans+=c(n0+n1-(num0+num1),n0-num0); 104 } 105 ans%=mod; 106 107 } 108 if(n1==1)//特判1的个数为1的时候 109 { 110 if(n0%2==1)//当0的个数为奇数是才会最终成为0 111 ans++; 112 } 113 } 114 else 115 { 116 for(i=1; i<=n0; i+=2)//奇数个0加上1个1时候，后面随便排列 117 { 118 num1=1; 119 num0=i; 120 if(n0+n1-(num0+num1)>0) 121 { 122 ans+=c(n0+n1-(num0+num1),n0-num0); 123 } 124 ans%=mod; 125 126 } 127 if(n1==1)//特判1的个数为1的时候 128 { 129 if(n0%2==0)//当0的个数为偶数是才会最终成为1 130 ans++; 131 } 132 } 133 printf("%I64d ",ans); 134 } 135 return 0; 136 }